The car is suspended beneath the helicopter by a steel cable. What is the lift force exerted on the air by the helicopter‘s rotor? (Hint: this will be the upward applied force acting on the helicopter.)
since the helicopter has to overcome gravity to accelerate. upward force - gravitational force net force F-up - mg ma F-up mg + ma F-up m (g +a) m 7500 + 1500 kg 9000 kg F-up 9000 (9.8 + 0.6) 9000 * 10.4 93600 N
F ma is your toddler right here. For the 1st one, how a lot weight might desire to the rotors raise? they might desire to lift the burden of the vehicle and the helicopter proper? So we upload them mutually. 5000kg + 1500kg 6500kg total that the rotors might desire to lift. all of us comprehend that the acceleration with the aid of gravity is 9.81m/s2, however the helicopter is accelerating upwards at 0.6m/s2. If its accelerating up, its making the burden sense like its heavier than it fairly is. So we upload the accelerations. 9.80 one + 0.6 10.41m/s2. Now in simple terms substitute into F ma F 6500kg * 10.41m/s2 67665 N For B, the stress interior the cable might in simple terms be the rigidity of the vehicle proper? Cuz the cable is in simple terms conserving the vehicle up. So returned F ma F 1500kg * 10.41m/s2 15615N
